Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - Review Exercises - Page 35: 32

Answer

$x= \frac{3}{2}-2t$ $y=1+2t$ $z=t$

Work Step by Step

Write the augmented matrix of the system of linear equations. $ \begin{bmatrix} 2 & 1 & 2 & 4\\ 2 & 2 &0 & 5\\ 2 & -1 & 6 & 2 \end{bmatrix} $ Divide the first row by 2. $ \begin{bmatrix} 1 & \frac{1}{2} & 1 & 2\\ 2 & 2 &0 & 5\\ 2 & -1 & 6 & 2 \end{bmatrix} $ Add -2 times the 1st row to the 2nd row to produce a new 2nd row. Add -2 times the 1st row to the 3rd row to produce a new 3rd row. $ \begin{bmatrix} 1 & \frac{1}{2} & 1 & 2\\ 0 & 1 &-2 & 1\\ 0 & -2 & 4 & -2 \end{bmatrix} $ Add 2 times the 2nd row to the 3rd row to produce a new 3rd row. $ \begin{bmatrix} 1 & \frac{1}{2} & 1 & 2\\ 0 & 1 &-2 & 1\\ 0 & 0 & 0 & 0 \end{bmatrix} $ Use the parameter t to represent z(where t is any real number) and substitute further for x and y. $z=t$ $y=1+2t$ $x= \frac{3}{2}-2t$
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